18 ANALYTICAL MECHANICS
represent the actions of these bodies by three force-vectors, W, T, and F Fig. 15, and then apply the conditions of equilibrium, betting equal to zero the sums of the components of the forces along the x- and y-axes, we get
SX ss F - T sin a = 0. (a)
£7 == _ff + T cos a = 0. (b)
Solving equations (a) and (b) we have
m __
W
and
cos a F = T sin a = W tan a.
W
DISCUSSION. — When a = 0, T=W
and F = 0. When a = |, !F = oo and
F=oo. Therefore no finite horizontal force can make the string perfectly horizontal. __
2. A uniform bar, of weight W and length a, is suspended in a horizontal position by two strings of equal length I. The lower ends of the strings are fastened to the ends of the bar and the upper ends to a peg. Find the tensile force in the strings.
The bar is acted upon by three bodies, namely the earth and the, two strings. We represent their actions by the forces W, Ti, and Ta, Fi&. 1 Oa. The tensile forces of the strings act at the ends of the bar. On the other hand the weight is distributed all along the rod. But we may consider it as acting at the middle point, as in Fig. 16a, or we may replace the rod
W
by two particles of weight — each, as shown in Fig. 16b. In the laat cane 2
the rigidity of the bar which prevents its ends from coming together IB represented by the forces F and — F.
Considering each particle separately and setting equal to zero the sums of the components of the forces along the axes, we obtain
1 OUD CC L \Jf
sin a - ~~ = 0,
for the first particle, and
2)7 i
sin a - = 0,